Question

Find the transformation of the equation 12x² −2y² +z² = 2xy if the origin is kept fixed and the axes are rotated in such a way that the direction ratios of the new axes are 1,−3,0; 3,1,0; 0,0,1.

Answer 1

The normalized axes are

$e'_1=\frac{1}{\sqrt{10}}(1,-3,0),e'_2=\frac{1}{\sqrt{10}}(3,1,0),e'_3=(0,0,1)$

or

$\begin{pmatrix} e'_1 \\ e'_2\\ e'_3 \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}}&0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}&0\\ 0&0&1 \end{pmatrix}\begin{pmatrix} e_1 \\ e_2\\ e_3 \end{pmatrix}$

- the basis in the rotated frame. The transformation of coordinates is given by

$\begin{pmatrix} x' \\ y'\\ z' \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}}&0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}&0\\ 0&0&1 \end{pmatrix}^{-1}\begin{pmatrix} x \\ y\\ z \end{pmatrix}$

So:

$\begin{pmatrix} x\\ y\\ z \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{10}} & \frac{-3}{\sqrt{10}}&0 \\ \frac{3}{\sqrt{10}} & \frac{1}{\sqrt{10}}&0\\ 0&0&1 \end{pmatrix}\begin{pmatrix} x' \\ y'\\ z' \end{pmatrix}$

$x = \begin{cases} x=\frac{x'}{\sqrt{10}}-\frac{3y'}{\sqrt{10}} \\ y=\frac{3x'}{\sqrt{10}}+\frac{y'}{\sqrt{10}}\\ z=z' \end{cases}$

$12x^2-2y^2+z^2=2xy$

$\frac{12(x'-3y')^2}{10}-\frac{2(3x'-y')^2}{10}+z'^2=\frac{2(x'-3y')(3x'-y')}{10}$

$12(x'^2-6x'y'+9'^2)-2(9x'^2+6x'y'+y'^2)+10z'^2=2(3x'^2-8x'y'-3y'^2)$

$-12x'^2-68x'y'+112y'^2+10z'^2=0$

Answer:

$5z'^2+56y'^2-6x'^2=34x'y'$